D+F 7.3.17

Let $R$ and $S$ be nonzero rings with identities $1_R$ and $1_S$. Let $\varphi:R \rightarrow S$ be a nonzero ring homomorphism.

(a) Prove that if $\varphi(1_R) \neq 1_S$ then $\varphi(1_R)$ is a zero divisor in $S$. Deduce that if $S$ is an integral domain, then every ring homomorphism from $R$ to $S$ sends the identity of $R$ to the identity of $S$.

(b) Prove that if $u$ is a unit in $R$ and $\varphi(1_R)=1_S$ then $\varphi(u)$ is a unit in $S$ and $\varphi(u^{-1})=\varphi(u)^{-1}$.

$\textit{Proof}$ :
(a): Let $\varphi(1_R)=y\in{S}-\{0\}$. (The identity in $R$ must be sent to a nonzero element of $S$ for $\varphi$ to be nonzero) $$y=\varphi(1_R)=\varphi(1_R1_R)=\varphi(1_R)\varphi(1_R)=y^2$$ $y=y^2$
$y-y^2=0$ (additive inverses exist because rings with respect to the addition operation are abelian groups)
$y(1_S-y)=0$ (distributive law for rings)
$ \therefore $ $y$ is a zero divisor.

If $S$ is an integral domain, then $S$ contains no zero divisors. So, in particular, $\varphi(1_R)$ is not a zero divisor. By the contrapositive of part (a), $\varphi(1_R) = 1_S$

(b) Assume that $u$ is a unit in $R$ and $\varphi(1_R)=1_S$. With $u$ being a unit of $R$, this means that
$\exists v \in{R}$ s.t. $uv=vu=1_R$. Since $\varphi$ is a ring homomorphism, $$\varphi(uv)=\varphi(vu)=\varphi(1_R)=1_S$$ $$\Longrightarrow \varphi(u)\varphi(v)=\varphi(v)\varphi(u)=1_S$$ Therefore, $ \varphi(u)$ is a unit in $S$.

We now want to show that $\varphi(u^{-1})=\varphi(u)^{-1}$. Note that $$1_S=\varphi(1_R)=\varphi(uu^{-1})=\varphi(u)\varphi(u^{-1})$$ $$\varphi(u)\varphi(u^{-1})=1_S$$ Since we have established that $\varphi(u)$ is a unit, it has an inverse. So, $$\varphi(u)^{-1}\varphi(u)\varphi(u^{-1})=\varphi(u)^{-1} \cdot 1_S$$ $$\therefore \quad \varphi(u^{-1})=\varphi(u)^{-1} \quad \blacksquare$$